Problem 36 - Calculus Volume 3

Question

Solve the initial-value problem. $$ 4 y^{\prime \prime}+4 y^{\prime}-8 y=0, \quad y(0)=2, \quad y^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The particular solution is \(y(t) = \frac{5}{3} e^t + \frac{1}{3} e^{-2t}\).

Step 1: Formulate the Characteristic Equation

Given the differential equation \(4 y'' + 4 y' - 8 y = 0\), we write the characteristic equation by substituting \(y = e^{rt}\), resulting in the equation \(4r^2 + 4r - 8 = 0\).

Step 2: Solve the Characteristic Equation

Factor or use the quadratic formula to solve \(4r^2 + 4r - 8 = 0\). The quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) gives roots: \(r = 1\) and \(r = -2\).

Step 3: Write the General Solution

The general solution to the differential equation is based on the roots of the characteristic equation. With distinct roots 1 and -2, the solution is \(y(t) = C_1 e^t + C_2 e^{-2t}\).

Step 4: Apply Initial Conditions

Use the initial conditions \(y(0) = 2\) and \(y'(0) = 1\) to solve for \(C_1\) and \(C_2\). For \(y(0) = 2\), substitute \(t = 0\) to get \(C_1 + C_2 = 2\). To use \(y'(0) = 1\), differentiate the general solution: \(y'(t) = C_1 e^t - 2C_2 e^{-2t}\) and substitute \(t = 0\) leading to \(C_1 - 2C_2 = 1\).

Step 5: Solve the System of Equations

Solve the linear system of equations: \(C_1 + C_2 = 2\) and \(C_1 - 2C_2 = 1\). From equation 1, \(C_1 = 2 - C_2\). Substitute into equation 2 to find \(2 - C_2 - 2C_2 = 1\), giving \(3C_2 = 1\), hence \(C_2 = \frac{1}{3}\). Substitute back to find \(C_1 = 2 - \frac{1}{3} = \frac{5}{3}\).

Step 6: Write the Particular Solution

Insert the values of \(C_1\) and \(C_2\) back into the general solution: \(y(t) = \frac{5}{3} e^t + \frac{1}{3} e^{-2t}\). This is the particular solution satisfying the initial conditions.

Key Concepts

Characteristic Equation

When you're dealing with linear differential equations, the characteristic equation is an essential tool. It originates from substituting a trial solution of the form \(y = e^{rt}\) into a differential equation. This helps to transform the problem from a differential equation into an algebraic equation. For our given problem, the differential equation is \(4 y'' + 4 y' - 8 y = 0\). By substituting the trial solution, we form the characteristic equation:

• \(4r^2 + 4r - 8 = 0\)

To solve this characteristic equation, you can use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this context, \(a = 4\), \(b = 4\), and \(c = -8\). Solving gives you the roots \(r = 1\) and \(r = -2\). These roots are vital as they indicate the nature of the general solution. When these roots are distinct, as they are here, they suggest that the solution involves linear combinations of exponential functions.

General Solution of Differential Equations

Once you have the roots of the characteristic equation, constructing the general solution becomes straightforward. For the problem at hand, we have distinct roots of \(r = 1\) and \(r = -2\). This indicates that the differential equation can be solved with a linear combination of terms involving these exponential roots. So, the general solution is:

• \(y(t) = C_1 e^t + C_2 e^{-2t}\)

Here, \(C_1\) and \(C_2\) are constants that will be determined using initial conditions. The solution reflects how the response of the system behaves over time, exemplified by exponential growth (from \(e^t\)) and decay (from \(e^{-2t}\)). It's crucial to understand that each different set of roots would yield different forms of solutions, such as repeated roots or complex roots, which would slightly alter the expression.

Initial Conditions in Differential Equations

Initial conditions are necessary to derive a unique solution from the general solution of a differential equation. In our task, we are given two initial conditions: \(y(0) = 2\) and \(y'(0) = 1\). These conditions are used to determine the specific constants \(C_1\) and \(C_2\) in the general solution. Here's the process:

• Substitute \(t = 0\) into the general solution \(y(t) = C_1 e^t + C_2 e^{-2t}\), giving \(C_1 + C_2 = 2\).
• Differentiate the general solution: \(y'(t) = C_1 e^t - 2C_2 e^{-2t}\) and substitute \(t = 0\), you find \(C_1 - 2C_2 = 1\).

Now you have a simple system of equations:

• Equation 1: \(C_1 + C_2 = 2\)
• Equation 2: \(C_1 - 2C_2 = 1\)

Solving these equations gives \(C_1 = \frac{5}{3}\) and \(C_2 = \frac{1}{3}\). By applying these conditions, you're narrowing down from the infinite solutions described by the general form to the one solution that aligns perfectly with the initial scenario presented in the problem.